All Answers 358087 Answers

The anwser is A   ........trapezoids have only one pair of sides that are parallel.....

I AN NOT GETTING WHAT YOU HAVE GIVEN PLEASE EXPLAIN YOUR ANS TO ME MORE BRIEFLY

(28+) - (27+) =

(28)(29)/ 2 - (27)(28)/2  =

(28) [ (29) - (27) ] / 2 =

(28/2) (2) = 28  =  (7)(8)/ 2 =   (7+)   ...so k = 7

If a quadrateral has exactly one pair of sides parallel it must be a:

A.trapezoid

B.rhombus

C.parallelogram

D.rectangle 

I do not believe that your point 2 is relevant.

Arc sine must be between -pi/2 and pi/2 

It is only neg in the 4th quad.

Cos(4th quad angle) is positive.

WolframAlpha agrees with me (according to Chris)

1+2+3+4+5+6+7+8+9+10 = 55  What is the Fastest way to do that?

$$\small{\text{$
1+2+3+4+5+6+7+8+9+10 =
\dbinom{n+1}{2} = \dbinom{10+1}{2} = \dbinom{11}{2}
= \dbinom{11}{9}=\dfrac{10\cdot 11}{2} = 5\cdot 11 = 55
$}}$$

According to Wikipedia......the cubit ranged from about 15 to 20.4 inches in length......if we consider the average man to be about 6 feet tall (60 in).....then  60/15  = about 4 cubits

$$\small{\text{$
\begin{array}{lrcl}
1. & \sin{(A)} &=& -\frac 35 \quad \Rightarrow \quad A = \arcsin{(-\frac 35 )} = -36.87\ensurement{^{\circ}} ~ (IV.)\\\\
2. & \sin{(180\ensurement{^{\circ}}-A)} &=& -\frac 35 \quad \Rightarrow \quad A = 180\ensurement{^{\circ}}- \arcsin{(-\frac 35 )} = 180\ensurement{^{\circ}} + 36.87\ensurement{^{\circ}} = 216.87\ensurement{^{\circ}} ~(III.)\\\\
\end{array}
$}}$$

And each sqrt has a plus and a minus

I'll agree, Melody...that, in general, we would have to consider it to be a fouth quad angle...but.....we would have to admit that, if it was a third quad angle, the sine would be negative and thus, as heureka's second answer indicates, the cosine would be negative there, as well........

Thanks Heureka

I'm still confused but I probably always will be :/

My recommendation????....practice a lot......

Also......on a test answer all the probllems you can do without "thinking" too hard......then, come back to the ones that give you trouble.......finally......

Realize that this isn't life or death.......it's just a test...not the end of the world.....

No it can't be in the 3rd quad

Inverse sine is confined to -pi/2 to pi/2

The general "formula" for the sum of the first "n" positive integers is given by:

n(n + 1) / 2   ...so we have.....for the first 10 integers....

(10)(11)/2   = 110/2  = 55

In terms of electromagnetism, one watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V).

And each rate of power is to multiply  with the hour you use it.

I think it has to be the 4th quadrant because of the restrictions on inverse sine

So cos must be positive.

But since the hypotenuse is 5 and the opp is 3 the adj must be 4 ( just like Heureka said)    so it is +4/5

What is cos[sin^-1(-3/5)]

$$\boxed{~\cos{[\arcsin{(x)}]} = \pm \sqrt{1-x^2}~}$$

$$\small{\text{$
\begin{array}{rcl}
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1-\left(-\dfrac 35 \right)^2}\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{1- \dfrac{9}{25} }\\\\
\cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} &=&
\pm \sqrt{\dfrac{16}{25} }\\\\
\mathbf{ \cos{\left[\arcsin{ \left( -\dfrac 35 \right) } \right]} } & \mahtbf{=} &
\mathbf{ \pm \dfrac{4}{5} }\\\\
\end{array}
$}}$$

Thanks for clearing that up Chris :))

Actually, Melody.....I think heureka meant to say that there are 1728 men in the mall.....if you add this to the 648 boys, you have my answer of 2376 males

You are very welcome.

Thanks for the feedback :))

Brilliant answer Melody, the working matches up perfectly with my notes and I can follow it clearly, thanks alot =)

Here's another way to tackle this one - although I still probably like geno's method better !!!!

Get y by itself in the first equation

x + y + 3z = 10      →   y =  10 - x - 3z

Substitute this into the second equation and simplify:

-4x + 2y +5z = 7    →   -4x + 2(10 - x - 3z) + 5z = 7   →   -4x + 20 -2x - 6z + 5z  = 7  →  -6x - z = -13  →

6x + z  = 13

So we have this system:

k*x + z = 3

6x + z = 13 

And notice that, if k =6, we would have this system:

6x + z  = 3      and

6x + z  = 13       but this is impossible, because two things added together can't possibly have two different outcomes 

So....as geno found, when k = 6, the system has no solution...!!!

$${\mathtt{\,-\,}}{\mathtt{11}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{12}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{15}}\right) = {\mathtt{16}}$$

Thanks Heureka    

But I do not understand the term kilowatt hours either.   

OR kilowatt for that matter.

What is a Watt?  I mean I know it has something to do with power but I still don't get it.    :(

The second term of a geometric sequence is 2 and the last term is 8. If the common ratio is 2√, how many geometric means are there between the first term and the last term?

$$\small{\text{$
\begin{array}{rcl}
t_2 &=& t_1\cdot r^1 \qquad | \qquad t_2 = 2 \quad \text{ and }\quad r = 2^{\frac 12} \\
2 &=& t_1 \cdot 2^{\frac 12}\\
t_1 &=& 2\cdot 2^{-\frac 12}\\
t_1 &=& 2^{ 1- \frac 12 }\\
\mathbf{t_1} & \mathbf{=} & \mathbf{2^{ \frac 12 }}\\\\\\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
t_n &=& t_1\cdot r^{n-1} \qquad | \qquad t_n = 2^3 \quad \text{ and }\quad t_1 = 2^{ \frac 12 } \quad \text{ and } \quad r = 2^{\frac 12}\\
2^3 &=& 2^{ \frac 12 } \cdot 2^{\frac 12 \cdot(n-1)} \qquad | \qquad 1^2\\
2^6 &=& 2^1 \cdot 2^{n-1}\\
2^6 &=& 2^{1+n-1}\\
2^6 &=& 2^n\\
6 &=& n\\
\mathbf{ n } & \mathbf{=} & \mathbf{6} \\
\end{array}
$}}$$

We have 6 terms, so between the first term and the 6th term are 4 terms.

$$\small{\text{$
\begin{array}{rcl}
t_1 &=& 2^{\frac 12} = \sqrt{2}\\
t_2 &=& 2^{\frac 22} = 2 \\
t_3 &=& 2^{\frac 32} = 2^{\frac 22}\cdot 2^{\frac 12} = 2\cdot \sqrt{2}\\
t_4 &=& 2^{\frac 42} = 2^2 = 4\\
t_5 &=& 2^{\frac 52} = 2^{\frac 42}\cdot 2^{\frac 12} = 2^2\cdot \sqrt{2}= 4\cdot \sqrt{2}\\
t_6 &=& 2^{\frac 62} = 2^3 = 8\\
\end{array}
$}}$$